chapter 5 : typical process...

CHAPTER 5 : TYPICAL PROCESS SYSTEMS

When I complete this chapter, I want to be able to do the following.

• Predict output for typical inputs for common dynamic systems

• Derive the dynamics for important structures of simple dynamic systems

• Recognize the strong effects on process dynamics caused by process structures

Outline of the lesson.

• Common simple dynamic systems- First order -Second order - Dead time - (Non) Self-regulatory

• Important structures of simple systems- Series - Parallel- Recycle - Staged

• Workshop


SIMPLE PROCESS SYSTEMS: 1st ORDER

The basic equation is:

)t(X K)t(Ydt

)t(dY=+τ

0 20 40 60 80 100 120

0.8

1

1.2

1.4

1.6

1.8

time

tank

con

cent

ratio

n

0 20 40 60 80 100 1200.5

1

1.5

2

time

inle

t con

cent

ratio

n

Maximumslope at“t=0”

Output changes immediately

Output is smooth, monotonic curve

At steady state

∆Y = K ∆X

τ

≈ 63% of steady-state ∆CA

∆X = Step in inlet variable

Would this be easy/difficult to

control?

K = s-s gain

τ = time constant

SIMPLE PROCESS SYSTEMS: 1st ORDER

These are simplefirst order systems

from severalengineeringdisciplines.

SIMPLE PROCESS SYSTEMS: 2nd ORDERThe basic equation is:

)()()(2)(2

22 tXKtY

dttdY

dttYd =++ ξττ

K = s-s gain , τ = time constant , ξ = damping factor

0 10 20 30 40 50 60 70 800

0.2

0.4

0.6

0.8

1

Time

Con

trolle

d V

aria

ble

0 10 20 30 40 50 60 70 800

0.2

0.4

0.6

0.8

1

Time

Man

ipul

ated

Var

iabl

e

0 20 40 60 80 100 120 140 160 180 2000

0.5

1

1.5

Time

Con

trolle

d V

aria

ble

0 20 40 60 80 100 120 140 160 180 2000

0.2

0.4

0.6

0.8

1

Time

Man

ipul

ated

Var

iabl

e

overdamped underdamped


control?

SIMPLE PROCESS SYSTEMS: 2nd ORDER

These are simplesecond

order systemsfrom severalengineeringdisciplines.

SIMPLE PROCESS SYSTEMS: DEAD TIME

time

Xin

Xout

θ = dead time

θ


control?

)()(

)()( sXesX

tXtX

ins

out

inoutθ

θ−=

−=

SIMPLE PROCESS SYSTEMS: INTEGRATOR

pump valve

Level sensorLevel sensor

Liquid-filledtank

Plants have many inventories whose flows in and out do not depend on the inventory (when we apply no control or manual correction).

These systems are often termed “pure integrators” because they integrate the difference between in and out flows.

outin F F dtdLA

dtdV ρρρρ −==

)()()()(LftFLftF

out

in

≠

≠


pump valve


Liquid-filledtank

outin FFdtdLA

dtdV ρρρρ −==

Fout

Fin

Plot the level for this scenario

time


pump valve


Liquid-filledtank

outin FFdtdLA

dtdV ρρρρ −==

Fout

Fin

time

Level


pump valve


Liquid-filledtank

• Non-self-regulatory variables tend to “drift” far from desired values.

• We must control these variables.

Let’s look aheadto when we apply control.

STRUCTURES OF PROCESS SYSTEMS

T

NON-INTERACTING SERIES

• The output from an element does not influence the input to the same element

• Common example is tanks in series with pumped flow between

• Block diagram as shown


T


• The output from an element does not influence the input to the same element

• Common example is tanks in series with pumped flow between

• Block diagram as shown

Gvalve(s) Gtank2(s)Gtank1(s) Gsensor(s)

v(s) F0(s) T1(s) T2(s) Tmeas(s)





)()()( sGsXsY

i

n

iΠ==1

In general:





)()()( sGsXsY

i

n

iΠ==1

In general:

With each element a first order system: )()(

)(11 +

Π== s

KsXsY

i

in

i τ





)()()( sGsXsY

i

n

iΠ==1

In general:

With each element a first order system: )()(

)(11 +

Π== s

KsXsY

i

in

i τ

• overall gain is product of gains

• no longer first order system

• slower than any single element

0 10 20 30 40 50 60 70-5

-4

-3

-2

-1

0

Time

Con

trolle

d Va

riabl

e

0 10 20 30 40 50 60 700

2

4

6

8

10

Time

Man

ipul

ated

Var

iabl

e

Step Response

• Looks as though some dead time occurs

• Smooth, monotonic, not first order

• Slower than any element

• K = Π (Ki)

STRUCTURES OF PROCESS SYSTEMSNON-INTERACTING SERIES

0.10/(5s+1) 1/(5s+1)-1.2/(5s+1) 3.5/(5s+1)






With each element a first order system with dead time:

)()()( i

11 +Π=

−

= seK

sXsY

i

sin

i τ

θ

Guidelines on step response

• Sigmoidal (“S”) shaped

• t63% ≈Σ (θi + τi) [not rigorous!]

• K = Π (Ki) [rigorous!]

• Usually, some “apparent dead time” occurs


Class Exercise: Sketch the step response for the system below.

τ = 2θ = 2

?


Class Exercise: Sketch the step response for the system below.

0 5 10 15 20 250

1

2

3

4

5DYNAMIC SIMULATION

Time

Con

trolle

d Va

riabl

e

0 5 10 15 20 250

1

2

3

4

5

Time

Man

ipul

ated

Var

iabl

e


Class Exercise: Sketch the step response for each of the systems below and compare the results.

Case 1

τ = 2 θ = 2 τ = 2 θ = 2

τ = 2 θ = 1 θ = 1τ = 2 & θ = 2

Case 2

0 2 4 6 8 10 12 14 16 18 200

1

2

3

4

time

case

1 re

spon

ses

0 2 4 6 8 10 12 14 16 18 200

1

2

3

4

time

case

2 re

spon

ses

Two plants can have different intermediate variables and have the same input-output behavior!

Step

Case1

Case2


PARALLEL STRUCTURES result from more than one causal path between the input and output. This can be a flow split, but it can be from other process relationships.

G1(s)

G2(s)

X(s) Y(s)

A → B → C

Example process systems Block diagram


PARALLEL STRUCTURES

G1(s)

G2(s)

X(s) Y(s)

))(()(

)()(

111

21

3

++

+=

sssK

sXsY p

τττ

If both elements are first order, the overall model is

Class exercise: Derive this transfer function


PARALLEL STRUCTURES can experience complex dynamics. Parameter is the “zero” in the transfer function.

G1(s)

G2(s)

X(s) Y(s)

0 1 2 3 4 5 6 7 8 9 10-0.5

0

0.5

1

1.5

time

outp

ut v

aria

ble,

Y’(t

)

0

-2

1

2

3

4

-1

Which wouldbe difficult/easy

to control?

Sample step response at t=0


PARALLEL STRUCTURE

Class exercise: Explain the dynamics of the outlet temperature after a change to the flow ratio, with the total flow rate constant.

T


0 5 10 15 20 2589

90

91

92

93

time

mix

ing

tem

pera

ture

0 5 10 15 20 250.4

0.5

0.6

0.7

time

fract

ion

by-p

ass

0 5 10 15 20 2589

90

91

92

93

time

seno

r out

put

T

Why an overshoot?

PARALLEL STRUCTURES: Explain the dynamics of the outlet temperature after a step change to the flow ratio.


PARALLEL STRUCTURE

Class exercise: Explain the dynamics of the outlet concentration after a step change to the solvent flow rate.

FA

CA0

V1CA1

V2CA2FS

CAS=0

reactant

solvent


0 10 20 30 40 50 600.05

0.06

0.07

0.08

0.09

0.1

time

solv

ent f

low

0 10 20 30 40 50 600.39

0.4

0.41

0.42

0.43

tank

2 c

once

ntra

tion

Why an inverse response?

PARALLEL STRUCTURE

Class exercise: Explain the dynamics of the outlet concentration after a step change to the solvent flow rate.


RECYCLE STRUCTURES result from recovery of material and energy. They are essential for profitable operation, but they strongly affect dynamics.

T0

T3T4

Process example Block diagram


RECYCLE STRUCTURES

GH1(s) GR(s)

GH2(s)

T0(s)

T1(s) T3(s)

T4(s)

T2(s)

)()()()(

)()(

sGsGsGsG

sTsT

HR

HR

2

1

104

−=


RECYCLE STRUCTURES

Class exercise: Determine the effect of recycle on the dynamics of a chemical reactor (faster or slower?).

T0

T3T4

• Exothermic reaction

• feed/effluent preheater

)/()(/ .)(/ .)(

1103300400

2

1

+==

=

ssGKKsGKKsG

R

H

H


Class exercise: Determine the effect of recycle on the dynamics of a chemical reactor (faster or slower?).

0 5 10 15 20 25 30 35 40 45 500

0.5

1

1.5

2

2.5

time

T4 w

ithou

t rec

ycle

T4 is a deviation variable

0 50 100 150 200 250 300 350 400 450 5000

5

10

15

20

25

time

T4 w

ith re

cycl

e

Without recycle, faster and smaller effect

With recycle, slower and larger effect

Different scales!


STAGED STRUCTURES

FR

FV

xB

xD

Tray n

Liquid

Liquid Vapor

Vapor

STRUCTURES OF PROCESS SYSTEMSSTAGED STRUCTURES

0 10 20 30 40 500.965

0.97

0.975

0.98

0.985

0.99

Time (min)

XD (m

olfra

c)

0 10 20 30 40 500.005

0.01

0.015

0.02

0.025

Time (min)

XB (m

olfra

c)

0 10 20 30 40 508530

8530.5

8531

8531.5

8532

8532.5

Time (min)

R (m

ol/m

in)

0 10 20 30 40 501.35

1.355

1.36

1.365

1.37x 104

Time (min)

V (m

ol/m

in)

Complex structure,smooth dynamics

“Steps” because analyzer provides new measurement only every 2 mintes.

OVERVIEW OF PROCESS SYSTEMS

Even simple elements can yield complex dynamics when combined in typical process structures.

We can

• Estimate the dynamicresponse based on elements and structure

• Recognize range of effects possible

• Apply analysismethods to yielddynamic model

CHAPTER 5: PROCESS SYSTEMS WORKSHOP 1

Four systems experienced an impulse input at t=2. Explain what you can learn about each system (dynamic model) from the figures below.

0 5 10 15 20 25 300

1

2

3

outp

ut

(a)

0 5 10 15 20 25 30-1

0

1

2

3

outp

ut

(b)

0 5 10 15 20 25 30-1

0

1

2

3

time

outp

ut

(c)

0 5 10 15 20 25 300

0.5

1

1.5

2

2.5

time

outp

ut

(d)


Using the guidelines in this chapter, sketch the response of the measured temperature below to a +5% step to the valve.

T

% openmin/m .

)s(v)s(F

)s(G valve30 10==

125021 3

0

1

+−

==s

min)/m/(K .)s(F)s(T)s(Gtank1

130001

1

2

+==

sKK

sTsTsG / .)()()(tank2 110

012

+=

=

sKK

sTsTsG measured

sensor

/ .

)()()(

(Time in seconds)


Sensors provide an estimate of the true process variable because the measurement is corrupted by errors.

• Discuss sources of noise in a measurement.

• Define the following terms for a sensor- Accuracy- Reproducibility

• Explain some process measurements needing (a) good accuracy and (b) good reproducibility

• Suggest an approach for operating a process when a key material property (composition, etc.) cannot be measured using an onstream analyzer.


We are designing the following reactor with recycle. We have two choices for the conversion in the reactor. Will the plant dynamics be affected by the selection?

Fresh feed flow is constant

Pure, unreacted feed

Pure product

X = 50%

X = 95%


When I complete this chapter, I want to be able to do the following.

• Predict output for typical inputs for common dynamic systems

• Derive the dynamics for important structures of simple dynamic systems

• Recognize the strong effects on process dynamics caused by process structures

Lot’s of improvement, but we need some more study!• Read the textbook• Review the notes, especially learning goals and workshop• Try out the self-study suggestions• Naturally, we’ll have an assignment!

CHAPTER 5: LEARNING RESOURCES

• SITE PC-EDUCATION WEB - Instrumentation Notes- Interactive Learning Module (Chapter 5)- Tutorials (Chapter 5)

• Software Laboratory- S_LOOP program

• Textbook- Chapter 18 on level modelling and control- Appendix I on parallel structures

CHAPTER 5: SUGGESTIONS FOR SELF-STUDY

1. Extend textbook Figure 5.1 for new input functions (additional rows): impulse and ramp.

2. Determine which of the systems in textbook Figure 5.3 can be underdamped.

3. Explain the shape of the amplitude ratio as frequency increases for each system in textbook Figure 5.1.

4. Discuss the similarity/dissimilarity between self regulation and feedback.

5. Explain textbook Figure 5.5.

6. Discuss the similarity between recycle and feedback.

CHAPTER 5: SUGGESTIONS FOR SELF-STUDY

7. Discuss how the dynamics of the typical process elements and structures would affect our ability to control a process. Think about driving an automobile with each of the dynamics between the steering wheel and the direction that the auto travels.

8. Formulate one question in each of three categories (T/F, multiple choice, and modelling) with solution and exchange them with friends in your study group.

chapter 5 : typical process...

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